∫_(−2) ^2 [x]dx=?


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Question Number 164929 by mathlove last updated on 23/Jan/22

$\underset{- 2}{\int^{2}} [x] d x = ?$
	Answered by TheSupreme last updated on 23/Jan/22
	$∫_a ^b ⌊x⌋dx=∫_a ^(⌊a⌋) ⌊x⌋dx+∫_(⌊a⌋) ^(⌊b⌋) ⌊x⌋dx+∫_(⌊b⌋) ^b ⌊x⌋dx= =⌊a⌋(⌊a⌋−a)+∫_(⌊a⌋) ^(⌊b⌋) ⌊x⌋dx+⌊b⌋(b−⌊b⌋) =−{a}⌊a⌋+∫...+⌊b⌋{b} ∫_(⌊a⌋) ^(⌊b⌋) ⌊x⌋=Σ_(i=⌊a⌋) ^(⌊b⌋−1) i=((⌊b−1⌋⌊b⌋)/2)−((⌊a⌋⌊a+1⌋)/2) ∫_(⌊a⌋) ^(⌊b⌋) ⌊x⌋dx=⌊b⌋{b}+((b(b−1))/2)−⌊a⌋{a}−((⌊a⌋(⌊a⌋+1))/2) ∫_(−2) ^2 ⌊x⌋dx=0+1−0−3=−2 you can see the integral as the sum of three parts. I) from a to ⌊a⌋ (negative area region) II) from ⌊a⌋ to ⌊b⌋ III) from ⌊b⌋ to b I and III are two rectangles of base: a−⌊a⌋ and b−⌊b⌋ and height: ⌊a⌋ and ⌊b⌋ II ⌊x⌋ is a step function. each step is rectangle that has base=1 and eight=i where i goes from ⌊a⌋ to ⌊b⌋ the last step ⌊b⌋−1<x<⌊b⌋ is tall ⌊b⌋−1 so we can conclude ∫_(⌊a⌋) ^(⌊b⌋) ⌊x⌋=Σ_(i=⌊a⌋) ^(⌊b⌋−1) i=((⌊b⌋⌊b−1⌋)/2)−((⌊a⌋⌊a+1⌋)/2)$
	$\int_{a}^{b} ⌊ x ⌋ d x = \int_{a}^{⌊ a ⌋} ⌊ x ⌋ d x + \int_{⌊ a ⌋}^{⌊ b ⌋} ⌊ x ⌋ d x + \int_{⌊ b ⌋}^{b} ⌊ x ⌋ d x =$ $= ⌊ a ⌋ (⌊ a ⌋ - a) + \int_{⌊ a ⌋}^{⌊ b ⌋} ⌊ x ⌋ d x + ⌊ b ⌋ (b - ⌊ b ⌋)$ $= - {a} ⌊ a ⌋ + \int . . . + ⌊ b ⌋ {b}$ $\int_{⌊ a ⌋}^{⌊ b ⌋} ⌊ x ⌋ = \underset{i = ⌊ a ⌋}{\sum^{⌊ b ⌋ - 1}} i = \frac{⌊ b - 1 ⌋ ⌊ b ⌋}{2} - \frac{⌊ a ⌋ ⌊ a + 1 ⌋}{2}$ $\int_{⌊ a ⌋}^{⌊ b ⌋} ⌊ x ⌋ d x = ⌊ b ⌋ {b} + \frac{b (b - 1)}{2} - ⌊ a ⌋ {a} - \frac{⌊ a ⌋ (⌊ a ⌋ + 1)}{2}$ $\int_{- 2}^{2} ⌊ x ⌋ d x = 0 + 1 - 0 - 3 = - 2$ $y o u c a n s e e t h e i n t e g r a l a s t h e s u m o f t h r e e p a r t s .$ $I) f r o m a t o ⌊ a ⌋ (n e g a t i v e a r e a r e g i o n)$ $I I) f r o m ⌊ a ⌋ t o ⌊ b ⌋$ $I I I) f r o m ⌊ b ⌋ t o b$ $I a n d I I I a r e t w o r e c t a n g l e s o f b a s e : a - ⌊ a ⌋ a n d b - ⌊ b ⌋ a n d h e i g h t : ⌊ a ⌋ a n d ⌊ b ⌋$ $I I ⌊ x ⌋ i s a s t e p f u n c t i o n .$ $e a c h s t e p i s r e c t a n g l e t h a t h a s b a s e = 1 a n d e i g h t = i$ $w h e r e i g o e s f r o m ⌊ a ⌋ t o ⌊ b ⌋$ $t h e l a s t s t e p ⌊ b ⌋ - 1 < x < ⌊ b ⌋ i s t a l l ⌊ b ⌋ - 1 s o w e c a n c o n c l u d e$ $\int_{⌊ a ⌋}^{⌊ b ⌋} ⌊ x ⌋ = \underset{i = ⌊ a ⌋}{\sum^{⌊ b ⌋ - 1}} i = \frac{⌊ b ⌋ ⌊ b - 1 ⌋}{2} - \frac{⌊ a ⌋ ⌊ a + 1 ⌋}{2}$
	Commented by mathlove last updated on 23/Jan/22

	$t h a n k s s i r$
	Answered by Ar Brandon last updated on 23/Jan/22

	$I = \int_{- 2}^{2} [x] d x = \underset{k = - 2}{\sum^{1}} \int_{k}^{k + 1} k d x = \underset{k = - 2}{\sum^{1}} k$ $= - 2 - 1 + 0 + 1 = - 2$
	Commented by mathlove last updated on 23/Jan/22

	$t h a n k s$
	Answered by TheSupreme last updated on 23/Jan/22
	$∫_0 ^x {t}dt=((⌊x⌋+{x}^2 )/2) ∫_0 ^x t−{t}=(x^2 /2)−(([x]+{x}^2 )/2)= =((x^2 −{x}^2 −[x])/2)=(([x](x+{x})−[x])/2)= =[x]((x+{x}−1)/2) =2+0−1−2−1=−2 if {x}=0 ⌊x⌋((x−1)/2)$
	$\int_{0}^{x} {t} d t = \frac{⌊ x ⌋ + {x}^{2}}{2}$ $\int_{0}^{x} t - {t} = \frac{x^{2}}{2} - \frac{[x] + {x}^{2}}{2} =$ $= \frac{x^{2} - {x}^{2} - [x]}{2} = \frac{[x] (x + {x}) - [x]}{2} =$ $= [x] \frac{x + {x} - 1}{2}$ $= 2 + 0 - 1 - 2 - 1 = - 2$ $i f {x} = 0$ $⌊ x ⌋ \frac{x - 1}{2}$
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